Material balance

calculations for jams

and marmalade

preparations

1

2

Contains

whole fruit or

fruit pieces:

JAMS

Contains fruit

pulps:

MARMALADES

No fruit piece,

made from

juice, clear

appearance;

JELLY

Difference between jams and

marmalades

Jam

Contains fruit pieces (whole or pieces).

Fruit is heated with sugar, acid and water to

activate the pectin (gelling agent) in the fruit.

If the fruit do not contain enough pectin and acid,

then pectin and acid (such as citric acid) are

added.

Whole berries (such as sour cherries) are most

frequently used without crushing.

Larger fruits such as apricots, peaches and

quinces are used after cutting into small pieces.

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PECTIN

4

Good jam has a soft even consistency, a bright

color, a good fruit flavor and a semi-jellied

texture that is easy to spread but has no free

liquid.

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Marmalade

In Europe, marmalade impies by heating

citrus fruit rind (most popularly oranges),

with pectin.

In Turkey, marmalade implies by heating

fruit pulps with pectin.

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Fruit spread

Refers to a jam or preserve with no added

sugar.

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Jelly

Obtained by heating the fruit juice with

pectin.

After initial heating, fruit pulp is filtered out.

Good jelly is clear and is tender (not hard),

enough to vibrate when moved.

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Other than these produts;

In the United States and European

countries, the following two products are

also produced.

Preserve: Obtained from whole fruit and

contains significant amount of fruit. (extra

geleneksel reçel)

Fruit spread: Preserve not containing

sugar.

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Example 3.21: The standard identity for jams and

preserves specifies that the ratio of fruit to added

sugar in the formulation is 45 parts fruit to 55 parts

sugar. A jam must have a soluble solid content of

at least 65% to produce a satisfactory gel. The

standard identity requires soluble solids of at least

65% for fruit preserves from apricot, peach, pear,

nectarine, plum, figs, and quince.

If the fruit contains 10% soluble solids and 100

grade pectin is used, calculate the weight of the

fruit, sugar, pectin and the amount of water

evaporated to produce 100 kg of fruit preserve.

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The process of making fruit preserves involves

mixing the fruit and sugar in the required ratio,

adding pectin and acid, and the concentrating the

mixture by boiling in a vacuum and atmospheric

pressure in a steam jacketed kettle until the

soluble solid content is at least 65%.

In soluble solid mass balance, the soluble

solids, acid content and sugar are considered

soluble solids; pectin excluded.

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Pectin grade: A 100 grade pectin is one

that will form a satisfactory gel at a ratio of

1 kg pectin to 100 kg sugar. A 150 grade

pectin is one that will form a satisfactory

gel at a ratio of 1 kg pectin to 150 kg

sugar.

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Mixing Preserrve,

100 kg

65% solids

Pectin (Z)

Sugar (Y)

Evaporatör

Water

Fruit, (X)

10% solids

Flow diagram for marmalade production

Answers

X = 49 kg fruit

Y = 60 kg sugar

Z = 0.6 kg pectin

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Example 3.22: The Turkish standard identity for

apricot marmalade specifies that the final product

should contain 40% apricot pulp (in essence “pure

fruit”). The marmalade of 150 kg with 68% soluble

solids will be produced from the fruit at 14% soluble

solids. To produce 150 kg of apricot marmalade,

calculate:

a) the weight of the apricot pulp, sugar, acid and

pectin,

b) the weight of the sugar and water to prepare pectin

solution (3%),

c) the amount of water to be evaporated in this

process.

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Data For the production of 150 kg marmalade, 600 g of

citric acid will be used, in addition to the acid

naturally present in the fruit. Citric acid will be

added at 50% aqueous solution.

Pectin of 150 grade will be used. The pectin used in

this process will be mixed with 5 times its weight of

sugar. The pectin will be added as 3% solution.

The glucose syrup will be added at a concentration

that the final product will contain glucose syrup at

5% of its weight. The soluble solid content of

glucose syrup will be 80%.

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Karıştırma

Marmalade,

150 kg

68% solids

Fruit, 60 kg

14% solids

Glucose syrup,

7.5 kg, 80% solids

Evaporatör

Water

Pektin

çözeltisi

hazırlama

Acid sol., 1.2 kg

50% solids

Water

(a)

Pectin

(Y)

Sugar

(X)

X2

X1

Flow diagram for marmalade production

Table 1 The recipe used for the production

of 150 kg of apricot marmalade

Ingredients Weight (kg)

Apricot pulp 60

Citric acid solution (50%) 1.2

Glucose syrup (80%) 7.5

Sugar 87

Pectin 0.58

Water 15.85

Total 172.13

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Result

The content of recipe should be cooked up

until 22.13 kg water is evaporated. After

evaporating 22.13 kg water, the

marmalade (150 kg) will contain 68%

soluble solids.

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Material balance calculations

for drying of foods

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Example 3.23

After drying of food initially containing 70%

moisture, 80% of its moisture content was

removed. Find out:

a) the moisture removed on the basis of 1 kg

of wet material.

b) The composition of dried food.

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22

Dryer Dried

apricots, (D)

Fresh apricots,

(W)

solids, 30%

water, 70%

Water, (X)

Flow diagram for dehydration of apricots

Answers

a) 0.56 kg moisture/1 kg wet material

b) 68.18% dried matter

31.82% moisture

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Example 3.24

How much weight reduction would result

when apricots are dried from 80% moisture

to 18% moisture?

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Example 3.25: 2080 kg/min of the hot air

left from a dryer with the absolute humidity

(mutlak nem) of H1 = 0.04 is mixed with

9135 kg/min of the air taken form

atmosphere with the absolute humidity of

H2 = 0.015. By this process, after heating

this mixture, the relatively hot and dry air is

sent back to the dryer. Find out the

absolute humidity of air mixture (H3).

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Absolute moisture is expressed as “kg of water

vapor/kg of dry air.” For example, if the abolute

moisture of air is 0.04, then this air contains 0.04

kg of water vapor per kg of dry air.

When the mass of wet air is given, this air is

composed of dry air and water vapor carried by

dry air. For example, if the absolute moisture

content of air is 0.04, then the mass of this air

mixture is 1.04 kg of dry air + water (1 kg of dry

air and 0.04 kg of water vapor).

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Mixing

Air mixture

Hot air,

H1= 0,04 kg of water/ kg of

dry air

Atmospheric air,

H2= 0,015 kg of water/ kg of dry air

Mixing of two air mixture with

different moisture content

Example 3.26: 2000 kg of fresh carrots

with 88% moisture are dried to the

moisture content of 4%. Hot air enters the

dryer with the absolute humidity of “0.01

kg water/kg dry air,” and then leaves the

dryer with the absolute humidity of “0.0225

kg water/kg dry air.” Find out the masses

of dried carrots leaving the dryer, and the

hot air entering the dryer as dry and wet

air.

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Dehydrator Dried carrots, (X)

Solids, 96%

Water, 4%

Fresh carrots, 2000 kg

Solids, 12%

Water, 88%

H2= 0,0225 kg water/kg DA

H1= 0,01 kg water/kg DA

Dring of carrots in dehydrator